The weight of in a 9-L cylinder at 27oC temperature and 16 atm pressure is (R = 0.0821 L atm K–1 mol — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

# Solution: Finding the mass of $CH_4$ **Step 1: Use the ideal gas law to find moles** $$PV = nRT$$ Given: - $P = 16$ atm - $V = 9$ L - $T = 27°C = 300$ K - $R = 0.0821$ L atm K$^{-1}$ mol$^{-1}$ **Step 2: Solve for n** $$n = \frac{PV}{RT} = \frac{16 \times 9}{0.0821 \times 300}$$ $$n = \frac{144}{24.63} = 5.85 \text{ mol} \approx 6 \text{ mol}$$ **Step 3: Convert moles to mass** Molar mass of $CH_4$ = 12 + 4(1) = 16 g/mol $$\text{Mass} = n \times M = 6 \times 16 = 96 \text{ g}$$ **Answer: The weight of $CH_4$ is approximately 96 g** (Option B) The calculation applies the ideal gas equation directly to find the number of moles, then multiplies by the molar mass of methane to get the final mass.